The following is a proposition about product sigma algebra from Folland's Real Analysis:

Proposition. If $A$ is countable, then $\otimes_{\alpha\in A}M_{\alpha}$ is is the $\sigma$-algebra generated by $\{\Pi_{\alpha\in A}E_\alpha:E_\alpha\in M_\alpha\}$.

Proof. If $E_{\alpha}\in M_{\alpha}$, then $\pi_\alpha^{-1}(E_\alpha)=\Pi_{\beta\in A}E_\beta$ where $E_\beta=X_\beta$ for $\beta\not=\alpha$; on the other hand, $\Pi_{\alpha\in A}E_\alpha=\cap_{\alpha\in A}\pi^{-1}_\alpha(E_\alpha)$. The result therefore follows from Lemma 1.1.

[Added]Lemma 1.1: If $\mathcal{E}\subset M(\mathcal{F})$ then $M(\mathcal{E})\subset M(\mathcal{F})$ where $M(\mathcal{E})$ denotes the sigma algebra generated by $\mathcal{E}$.

Where is the assumption "$A$ is countable" used in the proof?

  • 2
    $\begingroup$ What does Lemma 1.1 say? Presumably the point is that the intersection of sets in a $\sigma$-algebra is again in the $\sigma$-algebra, as long as the intersection is over only countably many sets. $\endgroup$ Aug 12, 2014 at 19:19
  • 1
    $\begingroup$ @AndresCaicedo: Thanks for your comment. I should have quoted Lemma 1.1, which would obviously not relate to the use of the "countability" assumption. $\endgroup$
    – user9464
    Aug 12, 2014 at 20:56

1 Answer 1


On the last line, we can only guarantee that the element $\cap_{\alpha \in A} \pi^{-1}(E_\alpha)$ is an element of $\otimes_{\alpha}M_\alpha$ if $A$ is countable: the $\sigma$-algebra is closed under countable intersections, not necessarily arbitrary ones.


You must log in to answer this question.